# Need Help Proving A Trig Identity Please Due Monday.

#### Deveno

MHF Hall of Honor
note that for ANY fractions a/b and c/d, saying a/b = c/d is THE SAME as saying: ad = bc.

so let's "cross-multiply" to eliminate the "fractions":

$$\displaystyle (\sin\theta - \cos\theta + 1)(\cos\theta) = (\sin\theta + \cos\theta - 1)(\sin\theta + 1)$$

a little algebra, now:

$$\displaystyle \sin\theta\cos\theta - \cos^2\theta + \cos\theta = \sin^2\theta + \sin\theta\cos\theta - \sin\theta + \sin\theta + \cos\theta - 1$$

$$\displaystyle \sin\theta\cos\theta - \cos^2\theta + \cos\theta = \sin^2\theta + \sin\theta\cos\theta + \cos\theta - 1$$

we have some terms that occur on both sides, so we can subtract them from each side:

$$\displaystyle -\cos^2\theta = \sin^2\theta - 1$$

(see it coming)? put the trig terms on one side, and the constant terms on the other:

$$\displaystyle 1 = \sin^2\theta + \cos^2\theta$$.

now, to be really slick, write this proof "upside-down", starting with the basic trig identity, and ending with what you wanted to prove.

#### Melcarthus

. It kind of funny I saw a problem similar to this one and a friend of mine said to cross multiply. So cross multiply to eliminate fractions. Then you foiled that left side of the equation and the right side. The two sin theta's canceled out ok. What happened to the cos theta minus 1? I understand the last part when you used the pythag ID.

#### Melcarthus

Another question, before the pythag id you have -cos to the second power0(theta)=Sin to the second power0(theta) minus 1. But how you get rid of second power to get to the equated side?